Finding the Greatest Number with Specific Remainders

The question asks for the greatest number that divides 126, 224, and 608, leaving remainders 2, 7, and 19, respectively. This type of problem involves understanding the relationship between division, divisors, and remainders, and applying the concept of the Greatest Common Divisor (GCD) or Highest Common Factor (HCF).

Understanding the Concept

If a number $d$ divides another number $a$ and leaves a remainder $r$, it means that $a$ can be written in the form $a = q \times d + r$, where $q$ is the quotient. Rearranging this equation, we get $a - r = q \times d$. This shows that $d$ must be a divisor of $a - r$.

In this problem, the greatest number we are looking for, let's call it $N$, must satisfy the following conditions:

• $N$ divides 126 leaving a remainder of 2. This means $N$ must divide $126 - 2$.
• $N$ divides 224 leaving a remainder of 7. This means $N$ must divide $224 - 7$.
• $N$ divides 608 leaving a remainder of 19. This means $N$ must divide $608 - 19$.

Therefore, the greatest number $N$ is the greatest common divisor (GCD) of the numbers obtained by subtracting the remainders from the original numbers.

Calculating the Modified Numbers

Let's subtract the given remainders from the respective numbers:

• $126 - 2 = 124$
• $224 - 7 = 217$
• $608 - 19 = 589$

The problem now reduces to finding the greatest common divisor (GCD) of 124, 217, and 589.

Finding the Greatest Common Divisor (GCD)

We can find the GCD by determining the prime factorization of each number.

• Prime factorization of 124:
  • Divide 124 by the smallest prime, 2: $124 \div 2 = 62$
  • Divide 62 by 2: $62 \div 2 = 31$
  • 31 is a prime number.
  • So, $124 = 2 \times 2 \times 31 = 2^2 \times 31$.
• Prime factorization of 217:
  • 217 is not divisible by 2 (it's odd).
  • Sum of digits $2+1+7=10$, not divisible by 3.
  • Doesn't end in 0 or 5, so not divisible by 5.
  • Try dividing by 7: $217 \div 7 = 31$.
  • 31 is a prime number.
  • So, $217 = 7 \times 31$.
• Prime factorization of 589:
  • 589 is not divisible by 2, 3, 5, 7.
  • Try dividing by 11: $589 \div 11$ is not a whole number.
  • Try dividing by 13: $589 \div 13$ is not a whole number.
  • Try dividing by 17: $589 \div 17$ is not a whole number.
  • Try dividing by 19: $589 \div 19 = 31$.
  • 31 is a prime number.
  • So, $589 = 19 \times 31$.

The prime factorizations are:

• $124 = 2^2 \times 31$
• $217 = 7 \times 31$
• $589 = 19 \times 31$

The greatest common divisor (GCD) is the product of the common prime factors raised to the lowest power they appear in any of the factorizations. In this case, the only common prime factor is 31, and it appears with a power of 1 in all factorizations.

Therefore, the GCD of 124, 217, and 589 is 31.

Conclusion

The greatest number that divides 126, 224, and 608 leaving remainders 2, 7, and 19, respectively, is 31.

The greatest common divisor of 124, 217, and 589 is 31.

Revision Table: Greatest Number and Remainders

Additional Information: GCD Concepts

The Greatest Common Divisor (GCD), also known as the Highest Common Factor (HCF), of two or more non-zero integers is the largest positive integer that divides each of the integers without leaving a remainder.

Methods to find GCD include:

• Prime Factorization Method: Find the prime factors of each number. The GCD is the product of all common prime factors raised to the lowest power they appear in any factorization.
• Euclidean Algorithm: For two numbers $a$ and $b$ ($a > b$), GCD($a, b$) = GCD($b$, $a \pmod{b}$), where $a \pmod{b}$ is the remainder when $a$ is divided by $b$. Repeat until the remainder is 0; the last non-zero remainder is the GCD. This can be extended to find the GCD of three or more numbers: GCD($a, b, c$) = GCD(GCD($a, b$), $c$).

In this problem, the prime factorization method was straightforward because 31 was a prominent factor.